Problem: $f(x)=x^6-x^3$ What is the coefficient for the term containing $(x+2)^4$ in the Taylor polynomial, centered at $x=-2$, of $f$ ? Choose all answers that apply: Choose all answers that apply: (Choice A) A $60$ (Choice B) B $1440$ (Choice C) C $240$ (Choice D) D $12$
Solution: Each term of a Taylor polynomial centered at $~x=-2~$ is in the form of $\frac{{{f}^{(n)}}(-2){{(x+2)}^{n}}}{n!}$. We need the fourth derivative of $~f\left( x \right)~$ evaluated at $~x=-2\,$. $f(x)=x^6-x^3$ $f'(x)=6{{x}^{5}}-3x^2$ $f''(x)=30x^4-6x$ $f'''(x)=120x^3-6$ $f''''(x)=360x^2$ So $f''''(-2)=1440\,$. Therefore, the coefficient of the term containing $~{{\left( x+2 \right)}^{4}}~$ is $\frac{1440}{4!}=60\,$.